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tangare [24]
3 years ago
12

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

ning nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus had a mass of 222 u, what speed does the alpha particle have when it is emitted?
Use conservation of momentum to solve.
Physics
1 answer:
alexandr1967 [171]3 years ago
4 0

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =  

where:  

M =222u is the mass of the original nucleus

v=420 m/s is the initial velocity of the nucleus

m_1 = 4 u is the mass of the alpha particle

v_1 is the final velocity of the alpha particle

m_2 = 222u-4u = 218 u is the mass of the daughter nucleus

v_2 = 350 m/s is the final velocity of the nucleus

Solving for v_1, we  find the final velocity of the alpha particle:

v_1 = \frac{Mu-m_2 v_2}{m_1}=\frac{(222)(420)-(218)(350)}{4}=4235 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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3 years ago
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

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Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
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The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
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Assume that negligible energy is lost when KE is converted into SE.
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0.06125k = 3.249
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